Chapter 4: Q133E (page 276)

4.133 Suppose xis a random variable best described by a uniform
probability distribution with c= 20 and d= 45.
a. Find f(x)
b. Find the mean and standard deviation of x.
c. Graph f (x) and locate $μ$ and the interval $μ \pm 2 σ$ onthe graph. Note that the probability that xassumes avalue within the interval $μ \pm 2 σ$ is equal to 1.

Short Answer

Expert verified

a. The probability density function is

$f (x) = \{\begin{cases} 0.04 & 20 \leq x \leq 45 \\ 0 & ; o t h e r w i s e \end{cases}$

b. The mean is 32.5 and the standard deviation is 7.2169

c. The lower limit is 18.0662 and the upper limit is 46.9338

Step by step solution

Given Information

Here x is a uniform random variable with parameters c=20 and d=45.

Finding the f (x)

The probability density function random variable x is given by

$f (x) = \frac{1}{d - c}; c < x < d$

Here, c=20 and d=45

So the pdf of x is:

$f (x) = \frac{1}{45 - 20} = \frac{1}{25} = 0.04$

Thus, f (x0 = 0.04 ; 20 < x <78C8 45

Finding the mean and standard deviation of x.

The mean of the random variable x is given by,

$μ = \frac{c + d}{2} = \frac{20 + 45}{2} = \frac{65}{2} = 32.5$

The standard deviation of x is given by,

$σ = \frac{d - c}{\sqrt{12}} = \frac{45 - 20}{\sqrt{12}} = \frac{25}{2 \sqrt{3}} = 7.2169$

Thus, the mean $μ = 32.5$ and standard deviation $σ = 7.2169$ .

The Graph

Here, the $2 σ$ limit is given by,

$μ - 2 σ = 32.5 - 2 \times 7.2169 = 18.0662 μ + 2 σ = 32.5 + 2 \times 7.2169 = 46.9338$

Here, the interval limits are outside of the actual range of random variable x.

The following graph depicts the relevant situation.

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	Mean	Standard Deviation
Tunnel Face	1.2	0.16
Tunnel Walls	1.4	0.2
Tunnel Crown	2.1	0.7

Short Answer

Step by step solution

Given Information

Finding the f (x)

Finding the mean and standard deviation of x.

The Graph

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